Find the angle at which the curves Gamma_1colon x=a cos

Find the angle at which the curves $\Gamma_1\colon x=a\cos\varphi,\ y=a\sin\varphi$ and $\Gamma_2\colon x=\dfrac{at^2}{1+t^2},\ y=\dfrac{at\sqrt3}{1+t^2}$ intersect.

Step 1: Identify both curves

$\Gamma_1$: $x^2+y^2=a^2$, the circle of radius $a$. For $\Gamma_2$, eliminating $t$ (using $t^2=x/(a-x)$ from the $x$-equation and $y=x\sqrt3/t$) gives

\[ y^{2} = 3ax-3x^{2} \;\Longrightarrow\; y^2+3\!\left(x-\frac a2\right)^2=\frac{3a^2}4, \tag{1} \] an ellipse centred at $\left(\dfrac a2,0\right)$.

Step 2: Intersection points

Substituting (1) into $x^2+y^2=a^2$: $2x^2-3ax+a^2=0 \Rightarrow (2x-a)(x-a)=0$. At $x=a$, $y=0$ is the point $t\to\infty$ (an asymptotic limit, not attained). At $x=a/2$: $y^2=3a^2/4 \Rightarrow y=\pm\dfrac{a\sqrt3}2$. Work at $P_1=\left(\dfrac a2,\dfrac{a\sqrt3}2\right)$.

Step 3: Slopes at $P_1$

Circle: $k_1=-x/y=-1/\sqrt3$. Ellipse, from (1): $2yy'=3a-6x \Rightarrow y'=\dfrac{3a-6x}{2y}$; at $P_1$, $3a-6x=3a-3a=0$, so $k_2=0$.

Step 4: Angle

\[ \tan\theta = \left|\frac{-1/\sqrt3-0}{1+0}\right| = \frac{1}{\sqrt3}. \tag{2} \]

Step 5: Verification

\[\text{At } P_2=\left(\tfrac a2,-\tfrac{a\sqrt3}2\right)\text{: } k_1=1/\sqrt3,\ k_2=0,\ \tan\theta=1/\sqrt3\ \checkmark\]